A takes 10 days less than the time taken by B to finish a piece of work. If both A and B can do it in 12 days, then the time taken by B alone to finish the work is = ?

Answer: Option A

Let B can alone finish the work = x days
So, A can alone finish the work = (x - 10) days
Now, one day work of A = $$\frac{1}{\text{x}-10}$$
and one day work of B = $$\frac{1}{\text{x}}$$
Now, given (A + B) can finish the work = 12 day
So, one day work of (A + B) = $$\frac{1}{12}$$
$$\begin{array}{rl}& \Rightarrow \frac{1}{\text{x}-10}+\frac{1}{\text{x}}=\frac{1}{12}\\ & \Rightarrow \frac{x+x-10}{x\times \left(x-10\right)}=\frac{1}{12}\\ & \Rightarrow \frac{2x-10}{x^2-10x}=\frac{1}{12}\\ & \Rightarrow 12\left(2x-10\right)=x^2-10x\\ & \Rightarrow 24x-120=x^2-10x\\ & \Rightarrow x^2-10x-24x+120=0\\ & \Rightarrow x^2-34x+120=0\\ & \Rightarrow x^2-30x-4x+120=0\\ & \Rightarrow x\left(x-30\right)-4\left(x-30\right)=0\\ & \Rightarrow \left(x-30\right)\times \left(x-4\right)=0\\ & \Rightarrow x=30,4\end{array}$$
if x = 4, then A alone can finish the work = 4 - 10 = -6, which is not possible.
So, x = 30
Hence, B can alone finish the work = 30 days